$\begingroup$So if I condition on the number of jumps to be n then there will be n terms in the product and since each jump is independent the expectation will become $(1+k)^n$ ? I know that I am supposed to get $\exp(\lambda k t)$ but how can I get that from $(1+k)^n$ ?$\endgroup$
– AndrewOct 8 '17 at 21:42

$\text{(1)}$ Note a subtlety here that got me momentarily confused: the product of $n$ i.d.d. random variables $\epsilon_1,\cdots,\epsilon_n$ with same distribution as $\epsilon$ is not the same as the $\text{n}^{\text{th}}$ power of variable $\epsilon$. Hence, you cannot directly collapse the product $\prod_{1\leq i \leq n}(1+\epsilon_i)$ to the power $(1+\epsilon)^n$, you first need to "inject" the expectation into the product.